Question

The correct order of the $1^{st}$ electron affinities of $N,O,F,S$ and $Cl$ is:

• A. $N<S<O<Cl<F$
• B. $Cl<S<N<O<F$
• C. $N<O<S<F<Cl$
• D. $S<Cl<N<O<F$

Answer 1

The correct option is C $N<O<S<F<Cl$

Electronic configuration of given species are as follows:
$N\left(7\right)=1s^{2}2s^{2}2p^3$
$O\left(8\right)=1s^{2}2s^{2}2p^4$
$F\left(9\right)=1s^{2}2s^{2}2p^5$
$S\left(16\right)=1s^{2}2s^{2}2p^{6}3s^{2}3p^4$
$Cl\left(17\right)=1s^{2}2s^{2}2p^{6}3s^{2}3p^5$
Electron affinity $\left(E_{ea}\right)$ of an atom or a molecule is defined as the amount of energy released or spent when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.
Along the period, from left to right $Z_{eff}$ increases, so $E_{ea}$ of the given period 2 elements will be N < O < F.
Cl has a more negative electron gain enthalpy than F because F has a high charge density and the added electron goes to the smaller energy shell (n=2) and hence, suffers a significant repulsion from the electrons already present in this shell. Thus $E_{ea}$: Cl > F
Similar reason for O and S. Oxygen being smaller in size than sulfur, has its valence shell electrons placed more closely than sulfur. Thus $E_{ea}$: S > O
After gaining an electron, F gets a completely filled electronic configuration. So $E_{ea}$ of F will be greater than O and S.
Therefore, overall order of $E_{ea}$ is:
$N<O<S<F<Cl$